Inverse Square Law as a special case of a Central Force
Both gravity, Fg = (-Gmm/r2)r, and electricity, Fe = (kqq/r2)r, are examples of a special
type of central force that has an inverse square law. Let’s apply the results of the
previous section on general central force to this special case.
To account for both gravity and electricity, we will use the symbol K to represent either
(–Gm1m2) or (kq1q2). Please note that K will be negative for gravity, while K can either
be positive (two like charges) or negative (two opposite charges) for electricity.
1. Potential Energy
Since we have a central force, we know that the force is conservative and hence we can
define a potential energy:
V(r) = - r-standardr F(r) dr = - r-standardr (K/r2) dr = -(-K/r) – (-K/rstandard) .
The usual choice for the standard position is infinity, so V(r) = K/r .
2. Effective Potential Energy
In the General Central Force, part 5, we defined an effective potential energy, Veff(r), as:
Veff(r) = V(r) + L2 /2mr2 ,
So in our case here of an inverse square central force, we have:
Veff(r) = K/r + L2/2mr2 .
so that the conservation of energy is essentially one-dimensional:
½ m r’2 + Veff(r) = E = constant .
3. Special case: stable equilibrium (circular motion)
In the General Central Force section, part 8, we noted that there will be a stable
equilibrium position where dVeff/dr = 0 if Veff is a minimum at that location. Let’s see if
and when the inverse square law has such a stable equilibrium position:
dVeff(r)/dr = d[K/r + L2/2mr2]/dr = -K/r2 – 2L2/2mr3 = 0 (for equilibrium)
Since r, L, and m are all positive, we get two cases:
Case a) If K>0, there is no positive value of r that satisfies the equation, so there is no
stable equilibrium radius, and hence no circular motion. This makes sense because K>0
implies a repulsive force.
Case b) If K<0, then there is a stable value of r: requilibirum = -L2/mK . Note that even
though there is a negative sign in this formula, the requillibrium is positive since K<0. It
makes sense that there would be a stable radius (circular motion), since K<0 implies an
attractive force which balances the “centrifugal force”, or in better words, the attractive
force causes the centripetal acceleration.
Let’s look at this circular motion from an energy standpoint.
E = ½ m r’2 + Veff(r)
but r’ = 0 for circular motion, so
E = Veff(requilibrium) = K/requilibrium + L2/2mrequilibrium2 .
Substituting in the requilibrium value from above (req = -L2/mK), we get
E = K/(-L2/mK) + L2/{2m[-L2/mK]2} = - mK2/L2 + mK2/2L2 = -mK2/2L2 .
Note that this energy is negative. That means the object is “bound” by the central force.
Also note that as L gets bigger, requilibrium gets bigger and the Energy gets closer to zero.
Recall that L = mvr = mr2’, or ’ = L/mr2 . Using requilibrium in this expression, we get:
’ = L/{m[-L2/mK]2} = K2m/L3 .
Also note that v = r’, so the circular speed in the orbit would be:
v = requilibrium ’ = (-L2/mK)*(K2m/L3) = -K/L ,
which says the orbital speed will be smaller as L gets larger (and requilibrium gets larger).
Note that L = r × p so that L = rmv , or vK/L = -K/rmv , or
v = SQRT[-K/mr]. As the radius gets bigger, the circular speed gets smaller. There is
just one speed for each radius for circular motion.
4. Finding r(t) for an Inverse Square Central Force – straightforward approach
Using the procedure in part 6 under the previous General Central Force section, we have:
½ mr’2 + Veff(r) = E
(where Veff(r) = V(r) + L2 /2mr2 )
dr/dt = r’ = [2(E-Veff(r))/m ]½
t
0
dt =
r
ro
{1 / [2(E-Veff(r))/m ]½ } dr .
For an inverse square law force, F = K/r2, we get (see part 2 above)
Veff(r) = K/r + L2/2mr2 .
t
0
t
t dt =
=
r
ro
r
ro
{1 / [2(E-Veff(r))/m ]½ }dr ,
or
{1 / [2(E-{ K/r + L2/2mr2} )/m ]½ }dr .
Multiplying top and bottom of the integrand by r gives
t
= (m/2)½ ror {r / [Er2- Kr - L2/2m ]½ }dr .
The above integral is of the general form:
 {r / [cr2+br+a]½ }dr = [cr2+br+a]½ /c - (b/2c)  {1 / [cr2+br+a]½ }dr
and
 {1 / [cr2+br+a]½ }dr = (-1/[-c]½ ) arcsin{ (2cr+b) / [4ac-b2]½ ) }
as long as c < 0 and (4ac-b2) < 0 .
Therefore, we get:
(2/m)½ t =
[Er2- Kr - L2/2m]½ /E + (K/2E)*{ (-1/(-E)½)*arcsin[(2Er-K)/[(-4EL2/2m)-K2]½ ] } .
This gives t(r), but it looks like it might be really hard to get a nice expression for r(t).
5. Find r() using the tricky approach (using u = 1/r).
From part 7 of the previous General Central Force section, we have:
(-m/L2u2)*F(1/u) + -u
= d2u/d2 .
For the inverse square law force, we have F = K/r2 = Ku2 , so we now get:
-mK/L2 - u = d2u/d2 ,
or
d2u/d2 + u = -mK/L2 = constant..
The inhomogeneous solution for u is simply:
uinhomogenous = -mK/L2 .
The homogeneous solution is simply:
uhomogeneous = A cos(-c) .
Note: when  = c , uhomogeneous is as big as it gets, and so, with r = 1/u, r is as small as it
gets. This means the when  = c , we have the position of closest approach.
Therefore, we get:
u() = -mK/L2 + A cos(-c),
1/r = -mK/L2 + A cos(-c) ,
or
where L = mvr = mr2’.
6. Initial Conditions
1/r = -mK/L2 + A cos(-c) ,
where L = mvr = mr2’.
To determine the constants L, A and c in the above expression for r(), we need to know
something about the orbit.
Usually we determine the constants A and c from the initial conditions on r, r’, , and ’.
Note that r’ = 0 only for circular motion. What should r’ be equal to? Note that r’ =
dr/dt. To find this, we can take the time derivative of our basic equation:
d[1/r]/dt = d[ -mK/L2 + A cos(-c)]/dt
or
2
d[1/r]/dr*dr/dt = d[ -mK/L + A cos(-c)]/d*d/dt
which becomes
[-1/r2]r’ = -[A sin(-c)]*’
or
r’ = [A sin(-c)]*r2’
and with mr2’ = L, we have
r’ = (AL/m)* sin(-c) .
If we know the initial radius, ro, and the initial angular speed, ’o, we can use the
expression L = mr2’ to find L. If we know the initial r, r’ and , we can use 1/r = mK/L2 + A cos(-c) and r’ = (AL/m)* sin(-c) to find A and c .
If we can determine the orbit, and from the orbit determine the area enclosed by the orbit,
S, we can then use the expression we developed in the General Central Force section, part
8:
S = (L/2m)T
where T is the period of orbit and S is the area of the orbit,
to determine the period of the orbit.
7. Overview of r() for inverse square law force
1/r = -mK/L2 + A cos(-o) .
Let’s first make a short-hand substitution: define B = -mK/L2 , so we have
1/r = B + A cos(-c) .
Let’s further define ε = A / │B│.
a) If K>0 (repulsive force), then B < 0 .
The radius, r, can’t be negative, therefore we must have A > │B│, which means ε > 1 .
The minimum radius, rmin occurs when cos(-c) is greatest. This means rmin occurs for
=c, and so
1/rmin = -mK/L2 + A .
The angle at which the radius goes to infinity occurs when B + A(cos-c) = 0. Note that
there will be two angles for which this occurs. Not all angles will result in positive
values for r which means that the orbit is not closed.
1/r = B + A cos(-o)
b) If K<0 (attractive force), then B>0 and then A can take on any value. We look at
four distinct cases:
b-1) A = 0; which implies that ε = 0, and further: 1/r = B. But this is the case where r
= constant, and so is the circular motion case: rcircular = 1/B = -L2/mK .
This is the same result we obtained in part 3 above.
b-2) A < B; which implies that ε < 1. Since A < B, 1/r never reaches zero, and so there
is a closed orbit. There is a minimum radius and a maximum radius:
1/rmin = B + A
which occurs when (-c) = 0 , and
1/rmax = B – A
which occurs when (-c) = 180o.
This turns out to be an ellipse (see the next section).
b-3) ) A = B; which implies that ε = 1. Since A=B, 1/r reaches zero (r approaches
infinity) only when -c = 180o . As in case b-2, there is a minimum radius when
1/rmin = B + A
which occurs when (-c) = 0o.
This turns out to be a parabola (see the next section).
b-4) ) A > B; which implies that ε > 1. Since A > B, r approaches infinity (1/r = 0) at
two angles such that B + A cos(-c) = 0. This turns out to be like case a above (for
attractive force), and this is a hyperbola (again, see the next section).